An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with
her arms stretched out is 2.48 kg m2. What is the increase in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.40 kg m2?
A) 0.957 JB) 0.902 JC) 0.870 JD) 0.750 JE) 0.690 J
A) 0.957 JB) 0.902 JC) 0.870 JD) 0.750 JE) 0.690 J
1 answer:
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Answer:
ω = 4.07 rad/s
Explanation:
By conservation of the energy:
W = ΔK
where
Solving for ω:
Answer:
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
The damping constant can be found by
Finally, the mass m can be found by
Where g is approximately 32 ft/s²
Therefore, the required differential equation is
The initial position is
The initial velocity is
Answer: Hello there!
We know this:
The distance between the cars at t= 0 is D.
car 2 has an initial velocity of v0 and no acceleration.
car 1 has no initial velocity and a acceleration of ax that starts at t = 0
then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.
Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:
position of car 1 + position of car 2 = D
and in this way we could ignore constants of integration :D
for the position of each car we integrate again:
P1(t) = (1/2)ax*t^2 and P2(t) = v0t
v0t + (1/2)ax*t^2 = D
v0t + (1/2)ax*t^2 - D = 0
now we can solve it for t using the Bhaskara equation.
that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).
Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:
where again, you need to replace the values of v0, D and ax.
Answer:
a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,
b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm
Explanation:
Given that:
γ= 9.5 kN/m³ = 9500N/m3
b = 6 inches = 0.1524 m
t = 0.0013 mm
d = 2 inches = 0.0508 m
n = 1750 rpm
L = 9 ft = 2.7432 m
Ks = 1.25
g = 9.81 m/s²
a)
b)
dip =